Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. The two light bulbs are connected in series across a 120-V line. Afterwards, the two light bulbs are connected in parallel across the 120-V line. (h) In which situation is there a greater total light output from both bulbs combined?

Video transcript

Hey everyone welcome back in this problem. We have two heating coils of resistance. R one equals 13.1. OEMs and R two is equal to 12. OEMs controlled by a circuit that either puts the coils in series or in parallel. The power source provides 110V and were asked to determine the connection that delivers more heat using both coils. Okay, so when we talk about delivering more heat in a circuit, what this means is that we have a higher power. Okay, So if we have a higher power output then we have more heat. So what's the power output? How can we calculate a power output? Well, the power output p it's going to be equal to the voltage B times the current. I now in our problem we're given B of 100 and 10 volts and were given information about our okay, we're not giving information about the current. I but we know through OEMs Law that we can relate V. I. And R. And so our power which is equal to V times I can also be written as V squared over the resistance are. Okay. Alright, so let's just go through and calculate what the power would be in series, what it would be in parallel and figure out which one's going to give us a higher output, which is going to mean more heat. Okay, so in series, what's our resistance going to be? Okay, well our resistance and we're gonna say R. S. Is just going to be the sum of the individual resistance is So we're gonna have resistance one that's resistance to Which is gonna give us a resistance of 13.1 OEMs less OEMs which equals 25.1 homes. Alright, so we have our resistance are here we actually know that the power source the V squared is gonna be the same in series or in parallel. Okay. So all we really need to find out is this are okay if we have a larger value of our Okay, larger r Then we're going to get a smaller power output. P because we're dividing by a larger number. Okay, so in series our resistance is going to be 25.1. OEMs. What about in parallel in parallel? How do we calculate our Okay, we know that the inverse Are going to sell. Okay, so we have one over the resistance in parallel it's going to be equal to one over R1. That's one over R. two. And if we multiply we find a common denominator. Okay? And then we take the inverse. We can write this as the resistance in parallel is equal to R one. R two over R one plus R two. So this tells us that the resistance is 13. times 12 times divided by R one plus R two. Okay. Which we know is 25.1 comes from the problem before or the calculation we did previously. Okay and serious. And this is going to give us a resistance in parallel of 6.27 ohms approximately. So you'll notice that in parallel can I? Our resistance is smaller than the resistance in series. Okay. In parallel we have 6.27 homes In series. We have 25.1 arms. What this means is that when we take the voltage squared divided by the resistance in parallel, this is actually going to be larger than when we do the same with the resistance in series. Okay. Because we're dividing by a smaller number, so the result is going to be larger and this tells us that our power output in parallel is going to be greater than our power output. And serious. Okay, Alright. So if we go back up to our answer choices, we see that we found that the power output in parallel is going to be higher. We have a higher power output. That means that we're going to have more heat. So the connection that delivers more heat using both coils is going to be be the parallel connection. Thanks everyone for watching. I hope this video helped see you in the next one